Answer
$a.\quad R=10. \quad $Interval of convergence:$\quad -8\lt x\lt 12$
$b.\quad $ Interval of absolute convergence:$\quad -8\lt x\lt 12$
$c.\quad $ No values of x for which the series converges conditionally.
Work Step by Step
(See text: "How to Test a Power Series for Convergence".)
$\text{Step 1.}$
Use the Ratio Test to find the interval where the series converges absolutely.
Ordinarily, $|x-a|\lt R\quad $ or $\quad a-R\lt x\lt a+R.$
$\begin{align*}
\displaystyle \frac{u_{n+1}}{u_{n}}&=\displaystyle \frac{(x-2)^{n+1}}{10^{n+1}}\div\frac{(x-2)^{n}}{10^{n}}\\
&=\displaystyle \frac{(x-2)^{n+1}}{10^{n+1}}\cdot\frac{10^{n}}{(x-2)^{n}}\\
&=\displaystyle \frac{x-2}{10}\\
\displaystyle \lim_{n\rightarrow\infty}|\frac{x-2}{10}|&\lt 1\displaystyle \\
\text{ We solve the inequality:}\\
\displaystyle \frac{|x-2|}{10}&\lt 1\displaystyle \\
|x-2|&\lt 10\\
-10&\lt x-2\lt 10\\
-8&\lt x\lt 12 \end{align*}.$
Determine the center and radius:
$a-R\lt x\lt a+R$
$2-10\lt x\lt 2+10$
$a=2,\quad $
the radius is $R=10,$
the interval of absolute convergence is $-8\lt x\lt 12$
$\text{Step 2.}$
If the interval of absolute convergence is finite,
test for convergence or divergence at each endpoint.
$ x=-8\quad \Rightarrow$
$\displaystyle \sum_{n=1}^{\infty}\frac{(-8-2)^{n}}{10^{n}}=\sum_{n=1}^{\infty}(-1)^{n}\qquad$ ... divergent series
$ x=12\quad \Rightarrow$
$\displaystyle \sum_{n=1}^{\infty}\frac{(12-2)^{n}}{10^{n}}=\sum_{n=1}^{\infty}(1)^{n}\qquad$ ... divergent series
$\text{Step 3.}$
If the interval of absolute convergence is $a-R\lt x\lt a+R$,
the series diverges for $|x-a|\gt R.$
So,
$a.\quad R=10. \quad $Interval of convergence:$\quad -8\lt x\lt 12$
$b.\quad $ Interval of absolute convergence:$\quad -8\lt x\lt 12$
$c.\quad $ No values of x for which the series converges conditionally.
